Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T, class T1>
inline int lower(const T& v, T1 x) {
  return lower_bound((v).begin(), (v).end(), x) - v.begin();
}
template <class T, class T1>
inline int upper(const T& v, T1 x) {
  return upper_bound((v).begin(), (v).end(), x) - v.begin();
}
template <class T>
inline T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
};
template <class T>
inline T pw(T a, T p, T M) {
  T result = 1;
  for (T sq = a; p; p >>= 1, sq = (sq * 1ll * sq) % M) {
    if (p & 1) {
      result = result * 1ll * sq % M;
    }
  }
  return result;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& a) {
  os << "(" << a.first << "," << a.second << ")";
  return os;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const vector<T1, T2>& a) {
  bool f = 0;
  os << "[";
  for (auto& x : a) {
    if (f) os << ",";
    f = 1;
    os << x;
  }
  os << "]";
  return os;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const set<T1, T2>& a) {
  bool f = 0;
  os << "{";
  for (auto& x : a) {
    if (f) os << ",";
    f = 1;
    os << x;
  }
  os << "}";
  return os;
}
template <class T1, class T2, class T3>
ostream& operator<<(ostream& os, const map<T1, T2, T3>& a) {
  bool f = 0;
  os << "{";
  for (auto& x : a) {
    if (f) os << ",";
    f = 1;
    os << x.first << ":" << x.second;
  }
  os << "}";
  return os;
}
struct _ {
  _() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.precision(30);
    cout << fixed;
  }
};
const int N = 100;
int a[N][N];
int main() {
  int n, m;
  cin >> n >> m;
  int ix = 0, jx = 0, vx = 1e9;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
      if (vx > a[i][j]) {
        ix = i;
        jx = j;
        vx = a[i][j];
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      a[i][j] -= vx;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (a[i][j] != a[i][jx] + a[ix][j]) {
        cout << -1;
        return 0;
      }
    }
  }
  int k = 0;
  for (int i = 0; i < n; i++) {
    k += a[i][jx] + (m > n ? vx : 0);
  }
  for (int j = 0; j < m; j++) {
    k += a[ix][j] + (m <= n ? vx : 0);
  }
  cout << k << endl;
  for (int i = 0; i < n; i++) {
    for (int cnt = 0; cnt < a[i][jx] + (m > n ? vx : 0); cnt++) {
      cout << "row " << i + 1 << endl;
    }
  }
  for (int j = 0; j < m; j++) {
    for (int cnt = 0; cnt < a[ix][j] + (m <= n ? vx : 0); cnt++) {
      cout << "col " << j + 1 << endl;
    }
  }
  return 0;
}